Find the volume of the solid obtained by rotating the region bounded
y = 16 x and y = 2 x^2 about y =0
Find the volume of the solid obtained by rotating the region bounded about the x-axis by y=4x^2, x =1, and y = 0
Find the volume of the solid generated by rotating the region bounded by
y = \sin^2(x) and y = 0
between 0 and pi about the x-axis
washers: v=∫[0,8] π(R^2-r^2) dx
where R=16x and r=2x^2
= 2621454/15 π
shells: v=∫[0,128] 2πrh dy
where r=y and h=√(y/2)-y/16
= 2621454/15 π
Can you set up the other integrals?
To find the volume of the solids obtained by rotating the given regions, we can use the method of cylindrical shells. Here's how to find the volume for each case:
1. Solid obtained by rotating the region bounded by y = 16x and y = 2x^2 about y = 0:
Step 1: First, let's find the points of intersection of the two curves.
Set y = 16x equal to y = 2x^2:
16x = 2x^2
2x^2 - 16x = 0
2x(x - 8) = 0
x = 0 or x = 8
So, the region bounded by y = 16x and y = 2x^2 is from x = 0 to x = 8.
Step 2: Now, let's find the equation for the curve of the solid when rotated about y = 0.
The equation of this curve is given by y = 16x - 2x^2 (since y = 16x - y = 2x^2).
Step 3: To find the volume, we integrate the area of each cylindrical shell from x = 0 to x = 8.
The volume is given by the formula:
V = 2π ∫[a,b] x(y(x)) * h(x) dx
Where x(y) is the equation for the curve, h(x) is the thickness of each shell, and [a,b] is the interval of integration.
In this case, x(y) = (y + sqrt(y^2 + 64y)) / 8, and h(x) = dx = dy / (dy/dx).
Since dy/dx = 16 - 4x, dy/dx = 0 at x = 4.
So, h(x) = dy / (dy/dx) = dy / (16 - 4x).
Now, we can evaluate the integral:
V = 2π ∫[0,8] [(y + sqrt(y^2 + 64y)) / 8] * [dy / (16 - 4x)] dx
Solving this integral will give us the volume of the solid.
2. Solid obtained by rotating the region bounded by y = 4x^2, x = 1, and y = 0 about the x-axis:
Step 1: The region bounded by y = 4x^2, x = 1, and y = 0 is a finite region. So, we don't need to find any points of intersection.
Step 2: The equation for the curve of the solid when rotated about the x-axis is y = 4x^2.
Step 3: Using the same formula as before, the volume is given by:
V = π ∫[a,b] x(y(x)) * h(x) dx
In this case, x(y) = √(y/4), and h(x) = dx = (dy/dx) / dy.
Since dy/dx = 8x, dy/dx = 0 at x = 0.
So, h(x) = (dy/dx) / dy = dx / (8x).
The integral to evaluate becomes:
V = π ∫[0,1] [√(y/4)] * [dx / (8x)] dx
Solve this integral to find the volume.
3. Solid generated by rotating the region bounded by y = sin^2(x) and y = 0 between 0 and π about the x-axis:
Step 1: The region bounded by y = sin^2(x) and y = 0 is also a finite region, so no need to find points of intersection.
Step 2: The equation for the curve of the solid when rotated about the x-axis is y = sin^2(x).
Step 3: Using the formula for volume, we have:
V = π ∫[a,b] x(y(x)) * h(x) dx
In this case, x(y) = arcsin(√y), and h(x) = dx = (dy/dx) / dy.
Since dy/dx = 2sin(x)cos(x), dy/dx = 0 at x = 0 and x = π/2.
So, h(x) = (dy/dx) / dy = dx / (2sin(x)cos(x)).
The integral to evaluate becomes:
V = π ∫[0,π] [arcsin(√y)] * [dx / (2sin(x)cos(x))] dx
Evaluate this integral to find the volume.
Remember to substitute the correct limits of integration and solve the integrals according to the specific problem to obtain the volumes of the respective solids.